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We have a uniform rectangular tray of dimensions 40x60, with weight 3. We place the coordinate axes in such a way that the bottom-left corner of tray is at the origin and the tray sides are parallel with the axes.

Suppose we have a quantity A of weight $2$ at the point $(20,10)$ and a quantity B of weight $5$ at $(50,20)$. Find the centre of gravity of the tray with the weights on it.

Attempted solution: Let $(x,y)$ be the centre of gravity of the system, and let $M_x$ and $M_y$be the sum of the moments around the $y$ and $x$-axes respectively.

To find $x$ I used the principle of moments to find $M_x = 2(20)+5(50)+3x = (2+3+5)x = 10x$, giving us $x = \frac{290}{7}$. Also $M_y = 2(10)+5(20) = (2+3+5)y = 10y$, giving us $y = \frac{120}{7}$. So the centre of gravity is $( \frac{290}{7} , \frac{120}{7} )$

However, my book gives the answer $(38,12)$. Can someone tell me where I went wrong?

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    $\begingroup$ Are you sure that the correct answer is $(38,12)$? I think the correct answer is $(38,18)$. $\endgroup$
    – lucas
    May 17, 2016 at 20:54
  • $\begingroup$ @lucas Can you explain why that answer is true? I think you are right (textbook is probably wrong), but it may be better, if you can, to explain how you got that result in a full answer. $\endgroup$
    – Cataline
    May 17, 2016 at 20:56
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    $\begingroup$ You have three masses: $3$ at $(20,30)$, $2$ at $(20,10)$ and $5$ at $(50,20)$. You has forgotten the first. And also your moment calculation is wrong. $\endgroup$
    – lucas
    May 17, 2016 at 20:57
  • $\begingroup$ @lucas I understand now. I mistakenly assumed the centre of gravity of the tray was the same as the system. Thank you. $\endgroup$
    – Cataline
    May 17, 2016 at 21:03
  • $\begingroup$ Your welcome:-) $\endgroup$
    – lucas
    May 17, 2016 at 21:03

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